$ D = \left[\begin{array}{rr}2 & 2 \\ 2 & 0\end{array}\right]$ $ B = \left[\begin{array}{rr}1 & 4 \\ 5 & -1\end{array}\right]$ What is $ D B$ ?
Because $ D$ has dimensions $(2\times2)$ and $ B$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D B = \left[\begin{array}{rr}{2} & {2} \\ {2} & {0}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{4} \\ {5} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ B$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ B$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ B$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{1}+{2}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{1}+{2}\cdot{5} & ? \\ {2}\cdot{1}+{0}\cdot{5} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{1}+{2}\cdot{5} & {2}\cdot\color{#DF0030}{4}+{2}\cdot\color{#DF0030}{-1} \\ {2}\cdot{1}+{0}\cdot{5} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{1}+{2}\cdot{5} & {2}\cdot\color{#DF0030}{4}+{2}\cdot\color{#DF0030}{-1} \\ {2}\cdot{1}+{0}\cdot{5} & {2}\cdot\color{#DF0030}{4}+{0}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 6 \\ 2 & 8\end{array}\right] $